The correct option is
A True
Refer to the figure.
segAD⊥segBC
segBE⊥segAC
segCF⊥segAB
Let p1 = length of seg AD
p2 = length of seg BE
p3 = length of seg CF
1) Now, Area of △ABC = △=12×b×h
∴△=12×p1×a
△=12×p2×b
△=12×p3×c
Thus, 12×p1×a=12×p2×b=12×p3×c=△
∴p1=2△a
p2=2△b
p3=2△c
Thus, cosAp1+cosBp2+cosCp3
=cosA(2△a)+cosB(2△b)+cosC(2△c)
=acosA2△+bcosB2△+ccosC2△
=acosA+bcosB+ccosC2△
=12△[asinAcosAsinA+bsinBcosBsinB+csinCcosCsinC]
=1△[(a2sinA)sinAcosA+(b2sinB)sinBcosB+(c2sinC)sinCcosC]
=1△[RsinAcosA+RsinBcosB+RsinCcosC]
where, R = radius of circumcircle of triangle.
=R2△[2sinAcosA+2sinBcosB+2sinCcosC]
=R2△[sin2A+sin2B+sin2C]
=R2△[2sin(2A+2B2)cos(2A−2B2)+sin2C]
=R2△[2sin(A+B)cos(A−B)+2sinC.cosC]
=R2△[2sin(π−C)cos(A−B)+2sinC.cosC]
=R2△[2sinCcos(A−B)+2sinC.cosC]
=R2△[2sinC(cos(A−B)+cosC)]
=R2△[2sinC(cos(A−B)+cos(π−(A+B)))]
=R2△[2sinC(cos(A−B)−cos(A+B))]
=R2△[2sinC(2sinAsinB)]
=R2△[4sinAsinBsinC] (1)
Now, △=abc4R
△=abc4(a2sinA)
∴△=sinA.abc2a
∴△=sinA.bc2
∴sinA=2△bc
Similarly, sinB=2△ac and sinC=2△ab
From equation (1),
=R2△[42△bc×2△ac×2△ab]
=16R△2a2b2c2
=16R△2(abc)2
=16R△2(4R△)2
=16R△216R2△2
=1R
2) LHS=bp1c+cp2a+ap3b
=b(2△)ac+c(2△)ab+a(2△)bc
=2△[b2+c2+a2abc]
=2(abc4R)[b2+c2+a2abc]
=a2+b2+c22R